线性代数基础，求解如图逆矩阵问题
2个回答
富港检测技术（东莞）有限公司_
2024-04-02 广告
正弦振动多用于找出产品设计或包装设计的脆弱点。看在哪一个具体频率点响应最大（共振点）；正弦振动在任一瞬间只包含一种频率的振动，而随机振动在任一瞬间包含频谱范围内的各种频率的振动。由于随机振动包含频谱内所有的频率，所以样品上的共振点会同时激发...
点击进入详情页
本回答由富港检测技术（东莞）有限公司_提供
展开全部
A =
[B O]
[O C]
其中 B =
[1 2]
[2 -1]
C =
[1 0]
[3 2]
B^(-1) = (-1/5)*
[-1 -2]
[-2 1]
=
[1/5 2/5]
[2/5 -1/5]
B^2 = 5E, B^4 = 25E
C^(-1) = (1/2)*
[ 2 0]
[-3 1]
=
[ 1 0]
[-3/2 1/2]
C^2 =
[1 0]
[9 4]
C^4 =
[1 0]
[45 16]
A^(-1) =
[B^(-1) O]
[O C^(-1)]
=
[1/5 2/5 0 0]
[2/5 -1/5 0 0]
[ 0 0 1 0]
[ 0 0 -3/2 1/2]
|A^T| = |A| = |B||C| = (-5)2 = -10
A^4 =
[B^4 O]
[O C^4]
[B O]
[O C]
其中 B =
[1 2]
[2 -1]
C =
[1 0]
[3 2]
B^(-1) = (-1/5)*
[-1 -2]
[-2 1]
=
[1/5 2/5]
[2/5 -1/5]
B^2 = 5E, B^4 = 25E
C^(-1) = (1/2)*
[ 2 0]
[-3 1]
=
[ 1 0]
[-3/2 1/2]
C^2 =
[1 0]
[9 4]
C^4 =
[1 0]
[45 16]
A^(-1) =
[B^(-1) O]
[O C^(-1)]
=
[1/5 2/5 0 0]
[2/5 -1/5 0 0]
[ 0 0 1 0]
[ 0 0 -3/2 1/2]
|A^T| = |A| = |B||C| = (-5)2 = -10
A^4 =
[B^4 O]
[O C^4]
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
